# If a finite set of ideals generates a ring, then so does any set of arbitrary powers of those ideals. [duplicate]

In Lang's Algebra, pg 95 (3rd Revised Ed.), he concludes a proof on the Chinese Remainder Theorem with:

In the same vein as above, we observe that if $\mathfrak{a_1},\dots,\mathfrak{a_n}$ are ideals of a ring $A$ such that $$\mathfrak{a_1}+ \dots + \mathfrak{a_n}=A,$$ and if $\mathit{v_1},\dots,\mathit{v_n}$ are positive integers, then $$\mathfrak{a_1}^\mathit{v_1}+\dots+\mathfrak{a_n}^\mathit{v_n}=A.$$

He states the proof is a trivial consequence of the CRT. However, I have been unable to find one that is satisfactory. I can see that there is a set of elements, one from each ideal which sum to 1. I am at a loss to see how we can find such a set that satisfies the above claim.

• This has appeared several times on this site, for example Comaximal ideals in a commutative ring. Therefore vote to close. The proof using CRT is easy, because it reduces to the case that only one ideal is nonzero. – Martin Brandenburg Mar 12 '13 at 22:49
• Your right. This should be closed. The use of radicals is slick and quick. I am not seeing how CRT reduces it to the case that only one ideal is nonzero, and this is not spelled out in the other post. – Rachmaninoff Mar 13 '13 at 18:00
• To clarify for anyone interested, the book claims that "the proof is trivial" not that it trivially follows from CRT (it does say that the claim is "in the same vein" as CRT though). It can be proved immediately, without CRT, using the same argument as in the accepted answer to the linked question but with $n$ variables rather than $2$. – Tom Oldfield Jul 2 '14 at 16:46